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PPT – Velocity and acceleration analysis analysis using vectors Problem 1 PowerPoint presentation | free to download - id: 850b1f-YzhkY

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Velocity and acceleration analysisanalysis using

vectorsProblem 1

Link 2 rotates at a constant angular velocity.

Determine the velocity and acceleration of points

C and D.

Solution 1 Velocity analysis for C

VC4VC3VC3A2VA2

VA2 1 rad/s2in 2in/s Normal to link 2

Hence direction and magnitude are both known for

VA2

VC3A2 is normal to link 3. Only direction is

known

VC4VC3 is horizontal. Only direction is known

Absolute velocity vectors VA2 and VC3 start from

Ov. Relative velocity vectors never start from

Ov.

A2

VA2 2in/s D, M

VC3A2 1.45 in/s (approx) D, M?

Ov

C3

VC4VC3 1in/s (approx) D, M?

VC3A2 1.45 in/s Hence w3 VD3A2

/AC 1.45/80.18 rad/s

Scale 1 in1 in/s

Solution 1 Acceleration analysis for C

aC4aC3aC3A2aA2anC3A2 atC3A2anA2atA2

atC3A2 Only Direction is known normal

to AC

aC4aC3 Direction is known - horizontal

anC3A2 w23 AC 0.1828 in/s2 0.26 in/s2

Direction- Radially inward along AC

aA2 w22 OA (1 rad/s)2 2 in2in/s2 Direction

radially inward along O2A

aC4aC3 1.48 in/s (approx) D, M?

Oa

C

The vector diagram

aA2 w22 OA (1 rad/s)2 2 in 2in/s2 D, M

atC3A2 1.38 in/s2 (approx) D, M?

atC3A2 1.38 in/s2 Hence a3 atC3A2

/AC 1.38/80.17 rad/s2

anC3A2 w23 AC 0.1828 in/s2 0.26 in/s2

(approx) D, M

Solution 1 Velocity analysis for D

VB5VB3VB3A2VA2 VB5D6VD6

VB3A2 w3AB VD3A2 (AB/AC) 0.18 rad/s4

in 0.72 in/s Normal to link 6.

VA2 1 rad/s2in 2in/s Normal to link 2

VD6 is normal to link 6. Only direction is known

VB5D6 is normal to link 5. Only direction is

known

Absolute velocity vectors VA2 and VC3 start from

Ov. Relative velocity vectors never start from

Ov.

VB5D6 1.3 in/s (approx) D, M?

VB3A2 0.72 in/s (approx) D, M

A

D

VD6 1.5 in/s Hence w5 VD6 /O6D 1.5/40.38

rad/s

VD6 1.5 in/s D, M?

VB5D6 1.3 in/s Hence w5 VB5D6 /BD 1.3/40.33

rad/s

VA2 2in/s D, M

B

Ov

Scale 1 in1 in/s

Solution 1 Acceleration analysis for D

aB5aB3aB3A2aA2 aB5D6aD6 anB5D6

atB5D6anD6atD6anB3A2 atB3A2anA2atA2

atD6 Only Direction is known normal to

O6D

anD6 w26 O6D (0.38 rad/s)2 4 in0.57

in/s2 Direction radially inward along O6D

anB5D6 w52BD 0.33240.44 in/s2 Direction

radially inward along BD

atB3A2 Only Direction is known normal

to AC

atB5D6 a3AB atC3A2(AB/AC) 1.384/80.69

in/s2 Direction normal to BD

anB3A2 w23 AB 0.1824 in/s2 0.13 in/s2

Direction- Radially inward along AC

aA2 w22 OA (1 rad/s)2 2 in2in/s2 Direction

radially inward along O2A

Solution 1 Acceleration analysis for D

aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6

atB5D6 anA2atA2 anB3A2 atB3A2

Since D cannot be located using the current

order of vector addition, locating B is also

difficult.

Where is B3 ?

Where is D6? Where does anB5D6 begin?

Oa

Scale 1 in 0.5 in/s2

Solution 1 Acceleration analysis for D

Oa

Change order of addition of vectors

atD6 4.1 in/s2 D, M ?

aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6

atB5D6 anA2 anB3A2 atB3A2atA2

aB5aB3aB3A2aA2 aB5D6aD6 anD6atD6anB5D6

atB5D6 anA2atA2 anB3A2 atB3A2

atB3A2 2.8 in/s2 D, M?

aD 4.05 in/s2

Scale 1 in 0.5 in/s2

B

D

Velocity and acceleration analysisanalysis using

vectorsProblem 2

Determine the linear acceleration of AP4 if

a20

Solution 2 Acceleration analysis for P

w2 Sense of rotation is found from direction of

VP2

Analyze available information

- VP2 10.0 in/sec. OPP0.95.
- Hence w2 VP2/OPP 10.53 rad/sec.
- Hence anP2 w22 OPP105.34 in/sec2
- a2 0.
- Hence atP2 0
- VP4P2 10.77 in/sec.
- Hence acP4P2 w2VP4P2 10.53 10.77113.38

in/sec2.

Vector equation

aQ4aP4aP3 aP4P2aP2asP4P2 acP4P2anP2atP2

Solution 2 Acceleration analysis for P

aP4 D, M?

asP4P2 D, M?

asP4P2 55

aP4 18

aP2anP2 (Since atP20) w22 OPP 105.34 D, M

acP4P2 2w2VP4P2 113.38 in/s2

aP4 Direction Along OP (OpPO)

acP4P2 2w2VP4P2 113.38 in/s2

asP4P2 Direction -Along AP

atP2 0

anP2 w22 OPP105.34 Radially inwards along OpP

(OpPO)

Velocity and acceleration analysisanalysis using

vectorsProblem 3

Link 2 rotates at a constant w2 1 rad/sec. q2

60o. Find VA4 and aA4 if O2A 2.4 cm

Solution 3 Velocity analysis for A

VA3A4 is along the slider in 4. Only direction

is known

VA3 w2O2A 1 rad/s2.4cm 2.4 cm/s Normal to

O2A

VB4 is along the slider in 1. Only direction is

known

VA4 VB4 (since 4 is translating only, all

points on 4 have same linear velocity)

VA3 w2O2A 1 rad/s2.4cm 2.4 cm/s

A3

VA3A4 1.2 cm/s D, M?

A4

Ov

VA42.08 cm/s D, M?

Solution 3 Acceleration analysis for A

aA2 aA3 anA2atA2 aA4asA3A4 acA3A4

Here we observe a link 3 which slides relative

to a moving link 4. Hence there should be a

Coriolis term

atA2 a2O2A 02.40 Hence aA2 anA2

aCorioliswcontaining linkVrelcontained link

aA4 aB4 ? Along O2B4

aCA3A4 2w4 VA3A4

But w4 0 as link 4 translates relative to the

ground. Hence acA3A4 0

asA4A3 ? Along A2B4

aA2 anA2 w22O2A 2.4 cm/s2 Radially inward

from A to O2

aA4 1.2 cm/sec2 D, M ?

Oa

A4

asA3A4 2.08 cm/s2 D, M ?

aA3 aA2 anA2 2.4 cm/sec2

A3

Problem 4

w2 1.000

1.0

1.631

3.56

4.41

Find the relative acceleration of the slider

block with respect to the curved slot in the

mechanism shown in figure. Assume w21 unit

angular velocity clockwise, a20, O2A1 unit

length. Use graphical method of acceleration

analysis using vectors.

Velocity diagram

tangent

OV

VA4

0.835 w40.5

A4

VA21

VA4 A3

1.016

A2 , A3

Acceleration diagram

OA

w42O4A0.41

a4O4A

A4

A2 , A3

2w4VA4A3 V2A4A3/4.41/

aA4A3,sliding

tangent

normal